Fraction Word Problems for Grade 6 Math Practice

Let's solve each problem step by step.

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Problem 1:

Stella goes for a long walk and walks \(\frac{2}{10}\) mile. She rested for some time and walked \(\frac{4}{10}\) mile. How much did Stella walk in total?

#### Solution:
To find the total distance Stella walked, we need to add the two fractions:

\[
\text{Total distance} = \frac{2}{10} + \frac{4}{10}
\]

Since the denominators are the same, we can simply add the numerators:

\[
\frac{2}{10} + \frac{4}{10} = \frac{2 + 4}{10} = \frac{6}{10}
\]

We can simplify \(\frac{6}{10}\) by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

\[
\frac{6}{10} = \frac{6 \div 2}{10 \div 2} = \frac{3}{5}
\]

Thus, the total distance Stella walked is:

\[
\boxed{\frac{3}{5}}
\]

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Problem 2:

Amelia walks \(\frac{8}{10}\) of a mile to her school. Ava walks \(\frac{7}{10}\) of a mile to school. How much more does Amelia walk than Ava, and by what distance?

#### Solution:
To find out how much more Amelia walks than Ava, we need to subtract the distance Ava walks from the distance Amelia walks:

\[
\text{Difference} = \frac{8}{10} - \frac{7}{10}
\]

Since the denominators are the same, we can subtract the numerators directly:

\[
\frac{8}{10} - \frac{7}{10} = \frac{8 - 7}{10} = \frac{1}{10}
\]

Thus, Amelia walks \(\frac{1}{10}\) mile more than Ava.

\[
\boxed{\frac{1}{10}}
\]

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Problem 3:

The new playground had \(\frac{1}{5}\)th of its space for football ground, \(\frac{3}{5}\)th space for track and field activities. How much space is left after this now?

#### Solution:
To find the remaining space, we first need to determine the total fraction of the playground that is already allocated. The space allocated is:

\[
\text{Allocated space} = \frac{1}{5} + \frac{3}{5}
\]

Since the denominators are the same, we can add the numerators:

\[
\frac{1}{5} + \frac{3}{5} = \frac{1 + 3}{5} = \frac{4}{5}
\]

The total space of the playground is 1 (or \(\frac{5}{5}\)). To find the remaining space, we subtract the allocated space from the total space:

\[
\text{Remaining space} = 1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{5 - 4}{5} = \frac{1}{5}
\]

Thus, the remaining space is:

\[
\boxed{\frac{1}{5}}
\]

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Problem 4:

An empty box weighs \(3 \frac{1}{3}\) kg. It is then filled with \(14 \frac{1}{3}\) kg of fruit. What is the weight of the box when it is full?

#### Solution:
First, convert the mixed numbers to improper fractions:

\[
3 \frac{1}{3} = 3 + \frac{1}{3} = \frac{9}{3} + \frac{1}{3} = \frac{10}{3}
\]
\[
14 \frac{1}{3} = 14 + \frac{1}{3} = \frac{42}{3} + \frac{1}{3} = \frac{43}{3}
\]

Now, add the weight of the empty box and the weight of the fruit:

\[
\text{Total weight} = \frac{10}{3} + \frac{43}{3}
\]

Since the denominators are the same, we can add the numerators:

\[
\frac{10}{3} + \frac{43}{3} = \frac{10 + 43}{3} = \frac{53}{3}
\]

Convert the improper fraction back to a mixed number:

\[
\frac{53}{3} = 17 \frac{2}{3}
\]

Thus, the weight of the box when it is full is:

\[
\boxed{17 \frac{2}{3}}
\]

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Problem 5:

A black table has a width of \(\frac{1}{3}\) meter, a brown table has a width of \(\frac{2}{5}\) meter, and the white table has a width of \(\frac{4}{5}\) meter. Which table has the longest width?

#### Solution:
To compare the widths of the tables, we need to find a common denominator for the fractions \(\frac{1}{3}\), \(\frac{2}{5}\), and \(\frac{4}{5}\). The least common multiple of 3 and 5 is 15. Convert each fraction to have a denominator of 15:

\[
\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15}
\]
\[
\frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15}
\]
\[
\frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}
\]

Now, compare the numerators:

\[
\frac{5}{15}, \frac{6}{15}, \frac{12}{15}
\]

The largest numerator is 12, so \(\frac{12}{15}\) is the largest fraction. Therefore, the white table has the longest width.

\[
\boxed{\text{White table}}
\]

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Final Answers:

1. \(\boxed{\frac{3}{5}}\)
2. \(\boxed{\frac{1}{10}}\)
3. \(\boxed{\frac{1}{5}}\)
4. \(\boxed{17 \frac{2}{3}}\)
5. \(\boxed{\text{White table}}\)